I am looking for functions and/or constants that when being integrated from minus infinity to infinity produce 1. I think the Dirac delta function is one example but perhaps there are some more? References on useful material is also greatly appreciated.

KennyTM: Thank you - the question remains what kind of integrals from -oo to oo give a finite nonzero answer?
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vonjdJul 28 '10 at 11:57

A good place to start would be functions that tend towards 0 as they approach both positive and negative infinity.
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Justin L.Jul 28 '10 at 12:07

@vonjd, this answer is accurate, but I want to emphasize one thing. It seems that all examples in this page are actually probability distributions. Though this is an interesting class of examples, these are by no means the only examples. For instance, the function $\frac{1}{\sqrt{2\pi}} (1+x) e^{-x^2}$ fits the requirements as well.
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SrivatsanJul 26 '11 at 20:33

Any function f(x) which integrates to 1 over any range [a,b] fits this bill, since we can define g(x) = f(x) on [a,b], and 0 everywhere else.

Even if you only want continuous functions, restricting ourselves above to f(x) where f(a) = f(b) = 0 still satisfies this.

If you want continuous functions strictly > 0 everywhere, these are known as probability distributions (continuous on [-∞, ∞]). A large list of such functions can be found here. A few more notable examples are:

One good example is the standard Gaussian distribution, $\phi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$. This is the most straightforward example of a continuous probability distribution function as mentioned by KennyTM above.

The function which is 1 on the interval [0;1], and 0 elsewhere, is a non-continuous probability distribution function. The function which is 3 on [0;1] and -1 on (1;3], and so on and on. What kind of answer do you want? What kind of properties do you want your functions to have?

There really are too many functions to list, since multiplying any function by a C^oo function with compact support and then applying Kenny's trick gives you an answer.

If you take any odd function $F$ differentiable on $\mathbb{R}$ and such that $F(x)\to l$ (with $l$ a nonzero real) for $x\to \infty$, then $f(x) = \frac{1}{2l}F'(x)$ statisfies your request. For example $F(x) = \operatorname{arctan}(x)$